Figure 2 shows a sketch of a triangle ABC - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 1

Now, we calculate the lengths of the sides:

1. Length of $\boldsymbol{|AB|}$: $|AB| = \sqrt{(2^2 + 3^2 + 1^2)} = \sqrt{4 + 9 + 1} = \sqrt{14}$

2. Length of $\boldsymbol{|AC|}$: $|AC| = \sqrt{(3^2 + (-6)^2 + 4^2)} = \sqrt{9 + 36 + 16} = \sqrt{61}$

3. Length of $\boldsymbol{|BC|}$: $|BC| = \sqrt{(1^2 + (-9)^2 + 3^2)} = \sqrt{1 + 81 + 9} = \sqrt{91}$

Using the cosine rule:

$\cos(\angle BAC) = \frac{|AB|^2 + |AC|^2 - |BC|^2}{2 |AB| |AC|}$

Substituting the lengths:

• $|AB|^2 = 14$
• $|AC|^2 = 61$
• $|BC|^2 = 91$

We have: $\cos(\angle BAC) = \frac{14 + 61 - 91}{2 \sqrt{14} \sqrt{61}} = \frac{-16}{2 \sqrt{14} \sqrt{61}} = \frac{-8}{\sqrt{14} \sqrt{61}}$

Now we can calculate $\angle BAC$:

Thus, $\angle BAC = \cos^{-1}\left(\frac{-8}{\sqrt{14} \sqrt{61}}\right)$

Upon calculation, this yields: $\angle BAC \approx 105.9^\circ$

This confirms that the angle $\angle BAC$ is indeed $105.9^\circ$ to one decimal place.